Suppose that $\omega$ and $\eta$ are two distinct primitive $n$-th roots of unity, where $n > 2 \in \mathbb{N}$ and $\omega \neq -\overline{\eta}$. I'm wondering how to prove, using only algebraic-like methods, that $\omega - \eta \notin \mathbb{Q}$. I think that an analysis work would do the job but I want to use Galois-theory properties and field theory to do this question. Can you suggest me how to do it? Also hints would be gladly accepted.

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    $\begingroup$ You need more conditions, as for example $\;1,\,-1\;$ are distinct $\;n\,$-th roots of unit for any even $\;n\;$ , yet $\;1-(-1)=2\in\Bbb Q\;$ ... $\endgroup$ – DonAntonio May 21 '18 at 15:54
  • $\begingroup$ True thing, I added $n>2$ and the fact that roots are primitive. $\endgroup$ – Alberto Andrenucci May 21 '18 at 15:55
  • $\begingroup$ I think if you let $\eta = -\overline{\omega}$ you still have a counterexample. $\endgroup$ – B. Goddard May 21 '18 at 15:56
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    $\begingroup$ It seems artificial not to use your proof "clearly an analysis work would do the job" - how would you prove it? $\endgroup$ – Dietrich Burde May 21 '18 at 16:03
  • $\begingroup$ Sure thing, I should have said that I think that an analysis work would do the job (we can analitically write down in terms of sines and cosines so i think we can actually do the computation). I didn't do the problem in that way but I'll think about it. $\endgroup$ – Alberto Andrenucci May 21 '18 at 16:12

Here's an entirely algebraic proof. However, I do include a variant using basic properties of $\Bbb C$ to derive equation $(**)$ because that's how I originally came up with it.

Wlog, your two primitive $n$-th roots of unity are $\omega$ and $\omega^r$ with $2\le r\le n-1$ and $gcd(r,n)=1$. Galois theory of cyclotomic fields tells us the Galois group of $\Bbb Q(\omega)|\Bbb Q$ consists of the automorphisms induced by $\omega \mapsto \omega^s$ for $gcd(s,n)=1$. So $$\omega^r-\omega \in \Bbb Q$$ is equivalent to $$\omega^{sr}-\omega^s = \omega^r-\omega \text{ for all } s \in \Bbb Z \text{ with } gcd(s,n)=1. \quad (*)$$

Now in particular we can choose $s=-1$ and get

$$\dfrac{\omega^{-r}-\omega^{-1}}{\omega^r-\omega} =1 ;$$

multiplying both sides with $\omega^{r+1}$ gives

$$-1 = \omega^{r+1} \quad(**).$$

Thus $n | 2(r+1)$. Now $n=r+1$ would contradict $(**)$, so $n=2r+2$, which is even, so $r$ is odd, so $$4|n \text{ and } r=\frac{1}{2}n-1 \quad (***).$$

Now setting in particular $s=r$ in $(*)$ and noting that, by $(***)$, $r^2 \equiv 1$ mod $n$, gives $\omega-\omega^r = \omega^r-\omega$, thus $\omega = \omega^r$, contradiction.

(Note also that your extra condition $n>2$ is vacuously true / redundant since you demand $\omega \neq \eta$. The condition $\omega \neq - \bar\eta$ is a bit weird because actually, $\omega = - \bar\eta$ is the only interesting case, namely it's equivalent to $(**)$. Actually, $\omega \neq - \bar\eta$ is equivalent to the much weaker $\omega -\eta \notin \Bbb R$ -- or in other words, $(**)$ is equivalent to $\omega = -\bar\eta$ is equivalent to $\omega -\eta \in \Bbb R$. See next paragraph.)

(Proof of $(**)$ with complex numbers:) Choosing any embedding of those unit roots into $\Bbb C$, their difference being $\in \Bbb Q$ means it's $\in \Bbb R$, hence their imaginary parts are equal. Because they both lie on the unit circle, this implies $$-1 = \omega \cdot \omega^r = \omega^{r+1} \quad(**)$$

The above proof avoids this, but of course the automorphism $\omega \mapsto \omega^{-1}$ "is" complex conjugation.

Finally note that if only $\omega$ is demanded to be primitive, one has the counterexample $\omega^2-\omega=-1$ for $n=6$. (Which brings up a different proof idea: If $\omega^r-\omega = a \in \Bbb Q\setminus \{ 0\}$, then the $n$-th cyclotomic polynomial has to divide $X^r-X-a$, which one can maybe lead to a contradiction too.)

  • $\begingroup$ Nice thoughts and nice answer! Thanks a lot! $\endgroup$ – Alberto Andrenucci May 23 '18 at 13:52

That's true for any two roots of an irreducible polynomial $P(x)\in \mathbb{Q}[x]$. If we had $x_1+r=x_2$, for $r \in \mathbb{Q}$, $r\ne 0$, then the polynomials $P(x)$ and $P(x+r)$ would have a common root $x_1$, so, being irreducible, must be the same, contradiction.

  • $\begingroup$ That's a neat, strong statement and proof (+1). Still I'm happy I went through the special case with special arguments, if only because that gave the extra info about when the difference is in $\Bbb R$. $\endgroup$ – Torsten Schoeneberg May 24 '18 at 1:47
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    $\begingroup$ Thank you! Yes, that is an interesting question. Btw, the quote from Pascal is nice. $\endgroup$ – Orest Bucicovschi May 24 '18 at 2:49

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